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<h3 class="heading"><span class="type">Paragraph</span></h3>
<p>According to the theorem, the second solution is given by</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
y_2=a y_1 \ln x+x^{r_2} \sum_{n=0}^{\infty} d_n x^n=a y_1 \ln x+\sum_{n=0}^{\infty} d_n x^{n-1/2} \quad \textrm{with}~d_0=1.
\end{equation*}
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<p class="continuation">Substituting this solution into the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
&amp;a \ln x [\underline{x^2 y_1^{\prime \prime}+x y_1^{\prime}+(x^2-1/4) y_1}]+2 a x y_1^{\prime}+\sum_{n=0}^{\infty} (n^2-n) d_n x^{n-1/2}+\sum_{n=0}^{\infty} d_n x^{n+3/2}=0,
\end{aligned}
\end{equation*}
</div>
<p class="continuation">where the term with underline is zero. We also have</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
\begin{aligned}
2 a x y_1^{\prime}&amp;=2 a x \left( \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1/2}}{(2n+1)!}   \right)^{\prime}\\
&amp;=2 a x \cdot \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1/2)  x^{2n-1/2}}{(2n+1)!}\\
&amp;=2 a \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1/2)  x^{2n+1/2}}{(2n+1)!}.
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Therefore,</p>
<div class="displaymath process-math" data-contains-math-knowls="">
\begin{equation*}
2 a \sum_{n=0}^{\infty} (-1)^n \frac{(2n+1/2)  x^{2n+1/2}}{(2n+1)!}+\sum_{n=0}^{\infty} (n^2-n) d_n x^{n-1/2}+\sum_{k=2}^{\infty} d_{k-2} x^{k-1/2}=0.
\end{equation*}
</div>
<p class="continuation">Multiplying both sides by the factor <span class="process-math">\(x^{1/2}\text{,}\)</span></p>
<span class="incontext"><a href="sec5_5.html#p-238" class="internal">in-context</a></span>
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